Question:
For a fixed positive integer $n$, if
\[
D=
\begin{vmatrix}
n! & (n+1)! & (n+2)!\\
(n+1)! & (n+2)! & (n+3)!\\
(n+2)! & (n+3)! & (n+4)!
\end{vmatrix},
\]
then
\[
\frac{D}{n!(n+1)!(n+2)!}
=
\]
For a fixed positive integer $n$, if
\[
D=
\begin{vmatrix}
n! & (n+1)! & (n+2)!\\
(n+1)! & (n+2)! & (n+3)!\\
(n+2)! & (n+3)! & (n+4)!
\end{vmatrix},
\]
then
\[
\frac{D}{n!(n+1)!(n+2)!}
=
\]
Show Hint
Factorials in determinants are usually simplified by factoring common row or column factors first.
Updated On: Jun 3, 2026
- $-4$
- $-2$
- $2$
- $4$
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Factor common terms from rows and columns of a determinant.
Step 2: Meaning
Extracting $n!,(n+1)!,(n+2)!$ from rows gives \[ \frac{D}{n!(n+1)!(n+2)!} = \begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 1 & n+2 & (n+2)(n+3)\\ 1 & n+3 & (n+3)(n+4) \end{vmatrix}. \]
Step 3: Analysis
Apply \[ R_2\to R_2-R_1,\qquad R_3\to R_3-R_2. \] The determinant simplifies to \[ \begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 0 & 1 & 2(n+2)\\ 0 & 0 & 2 \end{vmatrix}. \]
Step 4: Conclusion
Since the matrix is upper triangular, \[ \det=1\cdot1\cdot2=2. \]
Final Answer: (C)
Factor common terms from rows and columns of a determinant.
Step 2: Meaning
Extracting $n!,(n+1)!,(n+2)!$ from rows gives \[ \frac{D}{n!(n+1)!(n+2)!} = \begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 1 & n+2 & (n+2)(n+3)\\ 1 & n+3 & (n+3)(n+4) \end{vmatrix}. \]
Step 3: Analysis
Apply \[ R_2\to R_2-R_1,\qquad R_3\to R_3-R_2. \] The determinant simplifies to \[ \begin{vmatrix} 1 & n+1 & (n+1)(n+2)\\ 0 & 1 & 2(n+2)\\ 0 & 0 & 2 \end{vmatrix}. \]
Step 4: Conclusion
Since the matrix is upper triangular, \[ \det=1\cdot1\cdot2=2. \]
Final Answer: (C)
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